Question 23.4: It is required to design a cylinder for a hydraulic pressure......

Given N=1000 \mathrm{kN}, \frac{t}{r_i}=1.5, \sigma_{\mathrm{all}}=70 ~\mathrm{MPa}

From the load capacity, we can write

W=p_i \times\left(\frac{\pi}{4} D_i^2\right)

1000 \times 10^3=p_i \times\left(\frac{\pi}{4}\right) D_i^2

\therefore \quad p_i=\frac{10^6 \times 4}{\pi \times D_i^2}=\frac{4 \times 10^6}{\pi \times\left(2 r_i\right)^2}

=\frac{10^6}{\pi \times r_i^2} (23.15)

From the maximum shear stress criterion,

\tau_{\max }=\tau_{\text {all }}

\tau_{\text {all }}=p_i \frac{r_o^2}{r_o^2-r_i^2}

0.5 \times 70=\frac{10^6}{\pi r_i^2} \times\left(\frac{r_o^2}{r_o^2-r_i^2}\right) (23.16)

Given that

\frac{t}{r_i}=1.5 \text { or } \frac{r_o-r_i}{r_i}=1.5

\therefore \quad \frac{r_0}{r_i}=1.5+1=2.5

r_0=2.5 r_i (23.17)

From Eqs. (23.16) and (23.17)

0.5 \times 70=\frac{10^6}{\pi r_i^2} \times\left\{\frac{\left(2.5 r_i\right)^2}{\left(2.5 r_i\right)^2-r_i^2}\right\}

solving, we get

r_i=104.05 \mathrm{~mm}

=105 \mathrm{~mm}

r_o=2.5 \times r_i=2.5 \times 105=262.5 \mathrm{~mm}

=270 \mathrm{~mm}

\frac{t}{r_i}=\frac{270-105}{105}=1.57>1.5 \text {, Hence } \mathrm{OK} \text {. }